Flipping Coins: Sequences vs Independent Flips Here is a problem I thought of:

*

*Suppose I am watching someone flip a fair coin. Each flip is completely independent from the previous flip.

*I watch this person flip 3 consecutive heads.

*I interrupt this person and ask the following question: If the next flip results in a "head", I will buy you a slice of pizza. If the next flip results in a "tail", you will buy me a slice of pizza.

My Question: Who has the better odds of winning?
I wrote the following simulation using the R programming language. In this simulation, a "coin" is flipped many times ("1" = HEAD, "0" = TAILS). We then count the percentage of times HEAD-HEAD-HEAD-HEAD appears compared to HEAD-HEAD-HEAD-TAILS:
#load library
library(stringr)

#define number of flips
n <- 10000000 

#flip the coin many times
flips = sample(c(0,1), replace=TRUE, size=n)

#count the percent of times HEAD-HEAD-HEAD-HEAD appears 
str_count(paste(flips, collapse=""), '1111') / n

0.0333663

#count the percent tof times HEAD-HEAD-HEAD-TAIL appears
str_count(paste(flips, collapse=""), '1110') / n

0.062555

From the above analysis, it appears as if the person's luck runs out: after 3 HEADS, there is a 3.33% chance that the next flip will be a HEAD compared to a 6.25% chance the next flip will not be a HEAD (i.e. TAILS).
Thus, could we conclude: Even though the probability of each flip is independent from the previous flip, it becomes statistically more advantageous to observe a sequence of HEADS and then bet the next flip will be a TAILS? Thus, the longer the sequence of HEADS you observe, the stronger the probability becomes of the sequence "breaking"?
Thanks
 A: You are misled by a technical issue: it looks like the str_count command in R counts the number of non-overlapping matches it finds. Therefore it will not correctly count all the times that three flips of "heads" are followed by a fourth.
For example, in the sequence
11111101110

(where 1 represents "heads" and 0 represents "tails"), the str_count command will tell you that 1111 only appears once, but 1110 appears twice. But in fact, if you look at the five times that 111 appears in the string (including overlaps), then it is followed by another 1 three times, and by 0 only twice.
Over many coin-flips, if you count all instances of 1111 or 1110 and allow overlaps, you should see each one about $\frac1{16} = 0.0625$ of the time. So three "heads" are equally likely to be followed by "heads" or "tails".
(Note that this does not mean that after three "heads", there is a $0.0625$ probability of "heads" and a $0.0625$ probability of "tails": there is a $\frac12$ probability of each. The $0.0625$ counts the probability of all four flips happening.)
